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An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in \(\bar{x}=18.12 \mathrm{kgf} / \mathrm{cm}^{2}\) for the modified mortar \((m=40)\) and \(\bar{y}=16.87 \mathrm{kgf} / \mathrm{cm}^{2}\) for the unmodified mortar \((n=32)\). Let \(\mu_{1}\) and \(\mu_{2}\) be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. a. Assuming that \(\sigma_{1}=1.6\) and \(\sigma_{2}=1.4\), test \(H_{0}\) : \(\mu_{1}-\mu_{2}=0\) versus \(H_{a}: \mu_{1}-\mu_{2}>0\) at level 01 . b. Compute the probability of a type II error for the test of part (a) when \(\mu_{1}-\mu_{2}=1\). c. Suppose the investigator decided to use a level 05 test and wished \(\beta=.10\) when \(\mu_{1}-\mu_{2}=1\). If \(m=40\), what value of \(n\) is necessary? d. How would the analysis and conclusion of part (a) change if \(\sigma_{1}\) and \(\sigma_{2}\) were unknown but \(s_{1}=1.6\) and \(s_{2}=1.4 ?\)

Step by step solution

01

Understanding the Problem

We have two sets of sample data: one for modified mortar and one for unmodified mortar. The task is to compare the tension bond strength of both. We are given the sample means, sample sizes, and standard deviations for both samples, and we need to perform hypothesis testing and compute probabilities related to error types.

02

Setting Up Hypothesis for Part (a)

For part (a), we need to set up the null and alternative hypotheses. The null hypothesis ( H_0"): ∆ = 0, where the difference in true means (μ1 - μ2"). The alternative hypothesis ( H_a"): ∆ > 0, indicating the modified mortar has greater strength. We need to test this at a significance level of α = 0.01.

03

Performing Hypothesis Test for Part (a)

The test statistic for a known variance test is given by: \[ Z = \frac{(\bar{x} - \bar{y}) - 0}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}} \]Substitute in the values to calculate: \(\bar{x} = 18.12\), \(\bar{y} = 16.87\), \(m = 40\), \(n = 32\), \(\sigma_1 = 1.6\), \(\sigma_2 = 1.4\).Calculate the test statistic Z and compare it with the critical value from the Z-table for α = 0.01.

04

Calculating Probability of Type II Error for Part (b)

For Type II error, the true mean difference is considered to be 1. We need to calculate the power of the test and from that, deduce the probability of Type II error (β"). The formula is:\[ \beta = P\left(Z < Z_{\alpha} - \frac{1}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}}\right) \].Determine β" using this new condition considering the critical value calculated previously.

05

Adjusting Sample Size for Part (c)

To ensure β = 0.10" and maintain a significance level of α = 0.05", we need to use the formula for sample size derived from power analysis. The needed formula is:\[ n = \left(\frac{Z_{\alpha} + Z_{\beta}}{\Delta/\sqrt{\sigma_1^2/m + \sigma_2^2/x}}\right)^2 \].Solve this equation for \Delta = 1\", and given \alpha = 0.05 \", \beta = 0.10\", m = 40",

06

Considering Unknown Variances for Part (d)

When variances \sigma_1" and \sigma_2" are unknown, we switch from a Z-test to a t-test, using sample standard deviations \(s_1 = 1.6\) and \(s_2 = 1.4\).The test statistic is:\[ t = \frac{(\bar{x} - \bar{y}) - 0}{\sqrt{\frac{s_1^2}{m} + \frac{s_2^2}{n}}} \]With degrees of freedom calculated using the Welch-Satterthwaite equation. Calculate t and compare it to the critical value from the t-distribution table.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Type II Error

A Type II error occurs in hypothesis testing when we fail to reject the null hypothesis (H₀) when it is actually false. In simple terms, it's like saying there is no effect or difference when there really is one.
To understand this better, imagine you are a detective examining a crime scene. A Type II error would be similar to not finding any evidence even though the crime was committed.
This error is usually represented by the Greek letter β (beta), and occurs when the test lacks the power to detect an actual effect. Power is calculated as 1 - β and indicates how likely the test will catch a true effect. If β is small, power is high, meaning you're less likely to make a Type II error.

• Reducing β often involves increasing the sample size or using a more sensitive measurement method.
• In practice, researchers might choose an acceptable β level, like 0.1, to keep the risk manageable.

Sample Size Calculation

Calculating the correct sample size for a study is crucial to ensure reliable results in hypothesis testing. It helps strike the right balance between detecting true effects and avoiding errors.
Imagine you are baking cookies and want them to be just the right size; the sample size in an experiment is kind of like determining the ideal amount of ingredients to achieve your goal.
For hypothesis testing, especially when trying to balance Type I error ( α) and Type II error ( β), sample size plays a pivotal role.
A larger sample size generally increases the sensitivity of the test, making it easier to detect small differences between groups.
When calculating sample size, we usually need:

• The desired significance level ( α), often set at 0.05 or 0.01.
• The desired power (1 - β), i.e., the probability of correctly rejecting H₀, typically set at 0.8 or 0.9.
• The expected effect size, which is a measure of the magnitude of the difference being detected.

The formula for calculating sample size differs based on the test being used but often involves determining how many subjects or items are needed to observe a required level of precision for the hypothesis test.

t-Distribution

The is a probability distribution used when we are dealing with small sample sizes, and especially when the population standard deviation is unknown. This makes it very handy for real-life scenarios where such information is often lacking.
Picture the t-distribution as a slightly wider, more spread-out version of the normal distribution. As sample sizes increase, the t-distribution gets closer to a normal distribution. This spread helps account for the variability in estimating the standard deviation from small samples.
It's often used in conjunction with the in hypothesis testing. Specifically, it helps us find critical values when constructing confidence intervals or testing hypotheses. In testing scenarios, the calculated is compared against these critical values to make decisions about the hypotheses.

• The has heavier "tails," meaning it accounts for the greater likelihood of extreme values, which is common in small samples.
• Degrees of freedom (df) are key when using this distribution, calculated as df = n - 1, where n is the sample size.
• As n increases, the curve of the t-distribution becomes more like a normal distribution.

Understanding the t-distribution ensures that you make accurate inferences and avoid wrong conclusions, especially when working with small data sets.